External product in complex K-theory

For compact Hausdorff spaces $X$ and $Y$, define the external product by

$a⊠b={\mathrm{pr}}_X^{\ast }(a){\mathrm{pr}}_Y^{\ast }(b)\in K^0(X\times Y)\mathrm{.}$

Equivalently, $[E]⊠[F]$ is represented by ${\mathrm{pr}}_X^{\ast }E\otimes {\mathrm{pr}}_Y^{\ast }F$. Pullback, distributivity, and the ring structure make this a choice-free bilinear map

$K^0(X)\otimes K^0(Y)⟶K^0(X\times Y)$

that satisfies $(a⊠b)(a^{\mathrm{\prime }}⊠b^{\mathrm{\prime }})=a{a}^{\mathrm{\prime }}⊠b{b}^{\mathrm{\prime }}$.

Assume AC for the reduced clause. If $X$ and $Y$ are based and well-pointed, and $a\in {\tilde{K}}^0(X)$ and $b\in {\tilde{K}}^0(Y)$, then $a⊠b$ restricts to zero on $X\vee Y$. Exactness for

$X\vee Y⟶X\times Y⟶X\wedge Y$

therefore supplies a class in ${\tilde{K}}^0(X\wedge Y)$ whose pullback is $a⊠b$. It is unique: restriction to the wedge is surjective because the two projections extend any pair of reduced classes on its two summands, and the same projection argument after one reduced suspension makes ${\tilde{K}}^0(\mathrm{\Sigma }(X\times Y))⟶{\tilde{K}}^0(\mathrm{\Sigma }(X\vee Y))$ surjective. In the bi-infinite exact sequence this kills the connecting homomorphism preceding quotient pullback, so quotient pullback is injective. This unique class is also denoted $a⊠b$ and is the reduced external product.